Question 258181
16x-4x^2=0 We need to find what values of x make the denominator zero because that is where the vertical asymptotes are.
4x(4-x)=0
x=0, x=4 are the zeros.
There will be 2 vertical asymptotes at x=0 and x=4
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Ed
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{{{graph(500,500,-10,10,-10,10,(3x^2+9x-30)/(16x-4x^2))}}}